Problem: What is the value of $\sqrt{15 - 6\sqrt{6}} + \sqrt{15 + 6\sqrt{6}}$?
Explanation: Solution 1:
Let $x = \sqrt{15 - 6\sqrt{6}} + \sqrt{15 + 6\sqrt{6}}.$  Then \[x^2 = \left( \sqrt{15 - 6\sqrt{6}} \right)^2 + 2 \sqrt{15 - 6\sqrt{6}} \sqrt{15 + 6\sqrt{6}} + \left( \sqrt{15 + 6\sqrt{6}} \right)^2 \] We observe that $\left( 15 - 6\sqrt{6} \right)\left( 15 + 6\sqrt{6} \right) = 15^2 - \left(6\sqrt{6}\right)^2 = 225 - 216 = 9$ because of difference of squares.  So \[x^2 = \left( 15 - 6\sqrt{6} \right) + 2\sqrt{9} + \left( 15 + 6\sqrt{6} \right)\] The $6\sqrt{6}$ terms cancel, and so $x^2 = 36.$  Since $x$ must be positive, then $x = \boxed{6}$ and not $-6$.

Solution 2:
Let $a+b\sqrt{6} = \sqrt{15+6\sqrt{6}}$ for some $a$ and $b$. Squaring, we get $(a^2+6b^2) + 2ab\sqrt{6} = 15 + 6\sqrt{6}$. After some experimentation, we see this is true if $a=3$, $b=1$. So $\sqrt{15+6\sqrt{6}} = 3+\sqrt{6}$. Similarly, we find that $\sqrt{15-6\sqrt{6}} = 3-\sqrt{6}$. So $\sqrt{15-6\sqrt{6}} + \sqrt{15+6\sqrt{6}} = (3-\sqrt{6}) + (3+\sqrt{6}) = \boxed{6}$.